Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}2x+4y &= 7 \\ 4x+3y &= 8\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $3y = -4x+8$ Divide both sides by $3$ to isolate $y$ $y = {-\dfrac{4}{3}x + \dfrac{8}{3}}$ Substitute this expression for $y$ in the first equation. $2x+4({-\dfrac{4}{3}x + \dfrac{8}{3}}) = 7$ $2x - \dfrac{16}{3}x + \dfrac{32}{3} = 7$ Simplify by combining terms, then solve for $x$ $-\dfrac{10}{3}x + \dfrac{32}{3} = 7$ $-\dfrac{10}{3}x = -\dfrac{11}{3}$ $x = \dfrac{11}{10}$ Substitute $\dfrac{11}{10}$ for $x$ back into the top equation. $2( \dfrac{11}{10})+4y = 7$ $\dfrac{11}{5}+4y = 7$ $4y = \dfrac{24}{5}$ $y = \dfrac{6}{5}$ The solution is $\enspace x = \dfrac{11}{10}, \enspace y = \dfrac{6}{5}$.